Kinetic energy can be calculated by the equation

E = 1/2 mV^{2}

Assuming that the car weighs around 1 Ton or 1000 kg and travels at 100 km/h or 27.7 m/s, the equation will look like this:

= 1/2 (1000kg)(27.7 m/s)^{2}

= 385 802 J

or roughly 385 kJ

So each car is carrying 385 kJ of kinetic energy.

For you to bring that car to a stop, you will need to burn 385 kJ of kinetic energy by one of the following methods:

A. Be sensible and use your brakes (might take longer, but you'll survive)

B. Drive into a wall

C. Drive into an oncoming car of the same size and moving at the same velocity, but in the opposite direction

Either way, the car cannot disperse more energy than what it possessed in the first place.

If you have two cars, you will double the amount of energy coming into the equation, hence double the amount of energy dispersed if they collide head one, but exactly the same as two separate cars driving into walls.

However, if one car is driving double the speed into a wall:

E = 1/2 mV^{2}

= 1/2 (1000kg)(55.5 m/s)^{2}

= 1 540 125 J

or roughly 1 540 kJ and is therefore 4 times as much as the car driving at 100 km/h

So to summarize:

Two cars into wall or two cars head on at 100 km/h disperses the same amount of energy, for there is only the kinetic energy that is brought into the equation by each moving car.

One car driving at 200 km/h has 4 times the kinetic energy of a car at 100 km/h and can therefore disperse 4 times as much energy, whether against a wall or another car...